sand is being dropped onto a conical pile in such a way that the height of the pile is always twice the base radius.what is the rate of change of the volume of the pile with respect to the radius when the radius of the pile is 12 inches?

Answer 1

you need dV/dr
V = (1/3) pi r^2 (2r) = (2/3) pi r^3
dV/dr = (2/3)pi * 3 r^2
= 2 pi r^2
when r = 12
dV/dr = 288 pi